Hi Kristofer, Quick one:
--- En date de : Mer 15.6.11, Kristofer Munsterhjelm <km_el...@lavabit.com> a écrit : > I haven't tested FPC (since my reimplementation of JGA's > strategy ideas was done before I moved to a more modular > design for Quadelect), but as far as I recall, the really > standout systems as far as strategy was concerned, were > Condorcet-IRV hybrids (C//IRV, Smith,IRV, Smith//IRV, but > interestingly not Landau//IRV or Landau,IRV). Next to those, > at some distance, was a sorta-system that's like Ranked > Pairs, only in that winners are ranked above losers in the > sorted order and otherwise the order is fixed, so the system > isn't neutral. > ("Winners are ranked above losers" means that if A beat B, > A>B is reached before B>A.) > That method is a worst-case (I think) estimate for Ranked > Pairs-type methods. Making the sorting order depend on, say, > the winner's FPP score, could be interesting, but I don't > think that would be monotone (but hey, let's try to find > burial resistance before we find monotone burial resistance > :-) > > I also read a paper about trying to find the method that > would minimize burial opportunities. It reached the > conclusion that in the three candidate case, one should > elect the candidate that beat the FPP winner. The author > further said that he didn't know how to generalize it beyond > three candidates. > I think it was this one: > http://www.nhh.no/Admin/Public/DWSDownload.aspx?File=%2FFiles%2FFiler%2Finstitutter%2Ffor%2Fdp%2F2008%2F1108.pdf > "Condorcet Methods - When, Why and How? ". Ah, very interesting. I will try implementing "BPW" myself and see if I can get anything. > And from your other mail: > > > Also am I mistaken or is FPC with three candidates > identical to C//IRV? > > The candidate with the lowest penalty is going to be > the one who is > > defeated by the "FPL." If you were to eliminate the > FPL, the candidate > > with the best FPC score would beat the remaining > candidate. > > Hm, let's see. The three candidates are A, B, and C, and > they're in a cycle. Say that C is the candidate with the > least first place votes. In other words, f(A) + f(B) > > f(A) + f(C) and f(A) + f(B) > f(B) + f(C), since f(A) + > f(B) + f(C) sums to the number of voters, which is a > constant for each election. > > A's penalty is f(B) + f(C). > B's penalty is f(A) + f(C). > C's penalty if f(A) + f(B). er... Is this right? I thought your penalty in the 3c case would have to be just a single candidate's first preferences. I think I am probably right here. If you draw a triangle with ABC, you have two cycle possibilities. In both cycles IRV elects the winner between A and B. And in both cycles, that same candidate is beaten by the "FPL" C. Kevin Venzke ---- Election-Methods mailing list - see http://electorama.com/em for list info